In an estimation of bromine by Carius method, 1.6g of an organic compound gave 1.88g of AgBr. The mass percentage of bromine in the compound is (Atomic mass, Ag=108 , Br=80g mol^-1 )

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Correct answer: 50. (50) Mass of organic compound =1.6 gm Mass of AgBr=1.88 gm Moles of Br= Moles of AgBr= 1.88188=0.01 Mass of Br=0.01 80=0.80 gm \% of \% of Br= 0.80 1001.60=50 \%

$In an estimation of bromine by Carius method, 1.6 g of an organic compound gave 1.88 g of AgBr . The mass percentage of bromine in the compound is (Atomic mass, Ag = 108, Br = 80 g mol^{- 1})$

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